Question

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v₁. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v₂. If v₂ = x v₁, the value of x will be ________.

Answer 1

Correct Answer: 2

To solve this problem, we use the photoelectric effect's principle. According to Einstein's photoelectric equation: KE_max = hf - ϕ, where KE_max is the maximum kinetic energy of the emitted electrons, h is Planck's constant, f is the frequency of the incident light, and ϕ is the work function of the metal. Given conditions: 1. Frequency of the incident light = 2 × threshold frequency, resulting in speed v₁. 2. Frequency increased to 5 × threshold frequency, resulting in speed v₂. Assuming f₀ is the threshold frequency and m is the electron's mass, derive two scenarios: For v₁: h(2f₀) = (1/2)mv₁² + ϕ — (1). For v₂: h(5f₀) = (1/2)mv₂² + ϕ — (2). Subtract equation (1) from (2): h(5f₀) - h(2f₀) = (1/2)mv₂² - (1/2)mv₁². Simplify: h(3f₀) = (1/2)m(v₂²-v₁²) — (3). From (1): hf₀ = ϕ. Substituting, equation (3) becomes: (3ϕ) = (1/2)m(v₂²-v₁²). Solve for v₂: (2*3ϕ)/m = v₂²-v₁², hence v₂² = v₁² + (6ϕ/m). From equation (1): hf₀ = ϕ = (1/2)mv₁². Thus, v₂² = v₁² + 3v₁². Therefore, v₂ = 2v₁, leading to x = 2. The value of x is confirmed to be within the given range [2, 2].